Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y - 2}{6y + 48} \div \dfrac{y^2 - 7y}{y^2 + y - 56} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{y - 2}{6y + 48} \times \dfrac{y^2 + y - 56}{y^2 - 7y} $ First factor the quadratic. $n = \dfrac{y - 2}{6y + 48} \times \dfrac{(y + 8)(y - 7)}{y^2 - 7y} $ Then factor out any other terms. $n = \dfrac{y - 2}{6(y + 8)} \times \dfrac{(y + 8)(y - 7)}{y(y - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (y - 2) \times (y + 8)(y - 7) } { 6(y + 8) \times y(y - 7) } $ $n = \dfrac{ (y - 2)(y + 8)(y - 7)}{ 6y(y + 8)(y - 7)} $ Notice that $(y - 7)$ and $(y + 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ (y - 2)\cancel{(y + 8)}(y - 7)}{ 6y\cancel{(y + 8)}(y - 7)} $ We are dividing by $y + 8$ , so $y + 8 \neq 0$ Therefore, $y \neq -8$ $n = \dfrac{ (y - 2)\cancel{(y + 8)}\cancel{(y - 7)}}{ 6y\cancel{(y + 8)}\cancel{(y - 7)}} $ We are dividing by $y - 7$ , so $y - 7 \neq 0$ Therefore, $y \neq 7$ $n = \dfrac{y - 2}{6y} ; \space y \neq -8 ; \space y \neq 7 $